# How does a high pass filter work?

"The simplest high pass filter is a capacitor in resistor in series, where the voltage across the capacitor is used as output." Quote wikipedia
The picture of this simple high pass filter is found at: http://upload.wikimedia.org/wikipedia/commons/f/fe/High_pass_filter.svg.

I wonna get some things clear, because the circuit is drawn in a way that I am not used to. Which of the terminals are connected to the -pole of the voltage generator?

And more importantly, can someone give me an intuitive understanding of why low frequencies are not allowed to pass through. What property of the capacitor makes the current get damped for low frequencies?
If you switch the position of capacitor and resistor you get a low-pass filter. I don't see why switching these two would alter anything except the fact that you are now measuring the voltage across the capacitor. Isn't this the same as just switching the position of the voltmeter.

CWatters
Homework Helper
Gold Member
Wikipedia actually says..

http://en.wikipedia.org/wiki/High_pass_filter

The simple first-order electronic high-pass filter shown in Figure 1 is implemented by placing an input voltage across the series combination of a capacitor and a resistor and using the voltage across the resistor as an output.

Note "across the resistor" not "across the capacitor"

Which of the terminals are connected to the -pole of the voltage generator?

The source is connected to the terminals on the left. The output is taken from the terminals on the right.

Ideal capacitors block low frequencies (they are open circuit at DC) and pass high frequencies (short circuit at infinitely high frequencies).

Compare the circuit with a potential divider. eg what happens if you replace the capacitor with a variable resistor who's value depends on frequency.

oops yeh that voltage across the resistor.

What I don't understand is why the capacitor blocks high frequencies. I asked my teacher today and his answer was roughly: "Imagine a low frequency. That means that the capacitor will get charged a lot due to the low frequency causing a current the same way for a long time. This means that the voltage across the resistor will be small, which is understandable if you accept the above argument."
But then I asked him: "Wont the extra charge put into the capacitor help the voltage across the resistor when the voltage turns?" to which he didn't really have an answer other than to look at the differential equation - kind of unsatisfying if you want a good intuition for what is happening.
Can you clarify more for me what happens when we have a high frequency compared to a low frequency.

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CWatters
Homework Helper
Gold Member
What I don't understand is why the capacitor blocks high frequencies.

It doesn't. Capacitors pass high frequencies and block low frequencies.

Best advice I can give is to think how a capacitor is physically made. You have two parallel plates that are seperated (in some cases by air). There is no way for DC to cross the gap between the plates so to DC (Low frequencies) a capacitor looks like a very high value resistor (aka an open circuit).

It's harder to explain how high frequencies cross the plates but (at this level of understanding) if you could remember that high frequency radio waves travel through air quite well. So to high frequencies a capacitor looks like a low value resistor (aka short circuit).

Then if you want to understand how high pass or low pass circuits work just replace the capacitor with the appropriate open or short circuit. See diagram.

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• filter.bmp
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This all boils down to the fact that for a very short duration of time a capacitor works as a good conductor. The capacitor can absorb some small charge, so there is current flowing through the circuit. But as that happens, the capacitor gets charged, so it has progressively more "anti-voltage", impeding the current. If the frequency of the source is high, or the capacitor is large, it will have very little "anti-voltage" building up, so the current flowing through it will be large. Small capacitor, or low frequency - not much current.

This all boils down to the fact that for a very short duration of time a capacitor works as a good conductor. The capacitor can absorb some small charge, so there is current flowing through the circuit. But as that happens, the capacitor gets charged, so it has progressively more "anti-voltage", impeding the current. If the frequency of the source is high, or the capacitor is large, it will have very little "anti-voltage" building up, so the current flowing through it will be large. Small capacitor, or low frequency - not much current.

ahh yes, but this was exactly what I thought as well. But then what then when the voltage turns (for the voltage is AC). Then wouldn't your voltage gap across the capacitor at some point not start "helping" the potential drop across the resistor rather than work against it?
Please understand that I see perfectly well what you mean. That for low frequencies the capacitor gets charged up and such takes a great deal of the potential. But I don't see what hinders the thought that the reverse logic can be applied at some time during the cycle - i.e. that the voltage across the capacitor is opposite the voltage across the generator and then actually makes the voltage across the resistor even bigger.

Observe I am not talking about the voltage across the resistor. The resistor does not help us analyze the problem, it just gets in the way. Whatever voltage we have on it, it can be deduced from the current passing through the capacitor.

Keep in mind, that due to the capacitor, the source voltage and the current through the circuit are 90 degrees out of phase. When the source voltage is at its maximum, the current is zero; as the source voltage begins to drop (and eventually take on the opposite sign), the current begins to flow in the opposite direction. When the source voltage is zero, the current is at its maximum; then it begins to wane, becoming zero when the voltage is at its minimum, then the previous cycle repeats, but in the opposite direction.

hmm okay. I don't understand why you wonna leave the resistor out of this. After all that is the component we are measuring the voltage across.
Your statement about the current through the capacitor makes sense but I need some elaboration. Was this meant as an explanation as to why the capacitor can never have a potential difference across it that makes the voltage gap needed across the resistor bigger than the emf of the generator?

CWatters
Homework Helper
Gold Member
I think you are making things more complicated than they are.

The impedance of the resistor is R.
The impedance of the capacitor is Z (where z is proportional to 1/f.)

Then think of it like a potential divider circuit made from two resistors one value R and the other Z. The potential divider equation is...

Vout = Vin R/(Z+R)

If f=0 (eg DC) then Z is very large and Vout -> 0 (eg DC is blocked)

If f=large then Z ->0 and Vout = Vin (eg HF is passed).

At a certain frequency Z will equal R. Then vout = 0.5*Vin. This is the corner frequency.

NascentOxygen
Staff Emeritus
"The simplest high pass filter is a capacitor in resistor in series, where the voltage across the capacitor is used as output."
You are describing a form of potential divider, where one element is a capacitor and the other is a resistor. As you know, the impedance of an ideal resistor is a constant, equal to R Ohms regardless of frequency. But the impedance of a capacitor is frequency-dependent, with an inverse relationship, equaling 1/(2πf) Ohms. (This means "the higher the frequency, the lower the impedance.")

So, combining the two as a potential divider, you have one of the elements smoothly decreasing in Ohms—while the other stays fixed—as you increase the frequency of your input signal.

As you approach the upper limit (e.g., as f ⟶ ∞) the capacitor becomes like a short-circuit. So you can ask the question: as f ⟶ ∞ is the capacitor short-circuiting the input directly to output (i.e., a high-pass filter), or is the capacitor short-circuiting the output to ground (i.e., a low-pass filter)?

CWatters
Homework Helper
Gold Member
ahh yes, but this was exactly what I thought as well. But then what then when the voltage turns (for the voltage is AC). snip....

If you want to understand it at that level perhaps read this..

http://www.personal.soton.ac.uk/jav/soton/HELM/workbooks/workbook_12/12_6_cmplx_impedance.pdf

The standard representations for a.c. electronic signals are v = V0ejωt and i = I0ejωt where V0 is the (real) amplitude of the a.c. voltage and I0 is the (real) amplitude of the a.c. current and j = √−1.

snip

The current (i) into the capacitor is equal to the rate of change of the charge on the
capacitor i.e.

i = dq/dt = C dv/dt = jωCv.

Hence, for a capacitor, the impedance Zc = v/i = 1/jωC

Remember ω =2∏f so the impedance Zc is proportional to 1/f